Q. 67 - Calculus

Question

Prove that if the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$ and $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^{2k}$ have the same radius of convergence $p$, then $p$ is $0,1,$ or infinite.

Step-by-Step Solution

Answer

Ans:  Hence, the only solution to the equations $0,1 and \infty$

Step 1. Given information.

given,

     $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$ and $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^{2k}$

Step 2. Consider the power series ∑ k = 0 ∞   a k x k and ∑ k = 0 ∞   a k x 2 k has the same radius of convergence p

  Also, let us consider $b_k=a_k{x}^k,\text{ so }{b}_{k+1}=a_{k+1}{x}^{k+1}$ 

Apply the ratio test for absolute convergence in the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$, that is

   $\underset{k\to \mathrm{\infty }}{lim} \left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \mathrm{\infty }}{lim} \left|\frac{a_{k+1}}{a_k}x\right|$

So according to the ratio test for absolute convergence, the series will converge only when $\left|\frac{a_{k+1}}{a_k}x\right|<1$ 

Implies that  $\left|x\right|<\frac{a_k}{a_{k+1}}$

Where, $\frac{a_k}{a_{k+1}}$is the radius of convergence of the series  $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$

Since we have already considered  the radius of convergence of the series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^k$ is $p$, therefore, $\frac{a_k}{a_{k+1}}=p$

Step 3. Again, we apply the ratio test for absolute convergence in the power series &#8721; k = 0 &#8734; &#8202; a k x k , that is

 $\underset{k\to \mathrm{\infty }}{lim} \left|\frac{b_{k+1}}{b_k}\right|=\underset{k\to \mathrm{\infty }}{lim} \left|\frac{a_{k+1}}{a_k}{x}^2\right|$

So according to the ratio test for absolute convergence, the series will converge only  

    $\left|\frac{a_{k+1}}{a_k}{x}^2\right|<1$

Implies that $|x{|}^2<\frac{a_k}{a_{k+1}}$

Where, $\sqrt{\frac{a_k}{a_{k+1}}}$ is the radius of convergence of the power series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^{2k}$

Since we have already considered  the radius of convergence of the series $\sum _{k=0}^{\mathrm{\infty }} a_k{x}^{2k}$ is $p$,

therefore $\sqrt{\frac{a_k}{a_{k+1}}}=p$

Step 4. From the calculation of the radius of convergence for both the series, we can write p = p

Hence, the only solution to the equations $p=\sqrt{p}$are $0,1 and \infty$